Macmillan Learning Write the equation for the dissociation of the weak acid HBrO, in water. equation: HBrO, (aq) + H₂O(aq) = H₂O*(aq)BrO (aq) Incorrect
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- Calculation at the initial point of a complexation titration: What is the pSr value of a 50.00 mL solution of 0.01000 M Sr2+ (formation constant, Kf = 5.348 x 108) after the addition of 0.00 mL of 0.02000 M EDTA which was buffered to pH 11.00? Round your final answer to two decimal places to the right of the decimal point.pH of solution 14.00 12.00 10.00 8.00 6.00 First 4.00 equivalence point 2.00 0 First midpoint Second equivalence point Third midpoint pH = pKa=12.32 Third equivalence point HPO4(aq) + OH(aq) PO4(aq) + H2O(1) Second midpoint pH = pKa = 7.21 H2PO4 (aq) + OH(aq) HPO42 (aq) + H2O(1) Using the Henderson- Hasselback equation, show how to create 2L of a 0.1 M KPhos pH 7.5 buffer using K2HPO4 and KH2PO4. The chart to the left should help you understand what pKa to start with. Show your work. pH = pKa = 2.16 H3PO4(aq) + OH(aq) H2PO4(aq) + H2O(l) 25.0 50.0 75.0 100.0 Volume of NaOH added (mL)Calculation at the pre-equivalence point region of a complexation titration: What is the pSr value of a 50.00 mL solution of 0.01000 M Sr2+ (formation constant, Kf = 5.348 x 108) after the addition of 11.7 of 0.02000 M EDTA which was buffered to pH 11.00? Round your final answer to two places to the right of the decimal point.
- Find AG and AS for the experiment with Urea Standard State/when it melts AG: Mass of Urea (in grams) = 3.64 g Final volume of solution in graduated cylinder = 6.2 mL T= 298.15 K K = [(NH2)2CO]eq R = 0.008314 kJ/mol K AG = -RT In K [(NH2)2CO] = mol of (NH2)2CO/L of solution AS: AG(KJ/mol) = AH(kJ/mol) - TAS(kJ/mol K – times it by 1000) T= 298.15 K AG = ASBlood pH is maintained at a range of 7.4. The following set of equations representsthe reactions of the blood buffering system. The pk, for the dissociation ofH,COg into HOg and H* is 6.3.Carbonic AnhydraseCO2 + H2OH2CO3> HCO3- + H+(a) (Calculate the ratio [HCO3 ]/[H,COg] at physiological pH. Show work.(b)) An individual with diabetes presented at an urgent care with fatigue andnausea. Lab tests showed that their blood [HOg ]/[H,COg] ratio was 3.4. Whatwas the pH of the patient's blood? Show work.(c) ) Clinic staff also noticed that the patient was breathing rapidly. What isthe reason for this rapid breathing and how might it help alleviate the patient'scondition?Calculate the pH of a solution if [H3O+] = 3.4 x 10-2M Indicate whether the solution is acidic, basic, or neutral
- Utilising the provided class data generate the following graphs: I) Michaelis Menten; II) Lineweaver-Burk; and III) Hanes-Woolf. Ensure that you clearly label each graph,and add the relevant trendlines with equations. Table 1: Class data demonstrating the Absorbance at 700nm obtained for the alkaline phosphatase enzyme reaction Table 1 tube Abs700mm 1 0.000 2 0.060 2 0.090 4 0.140 5 0.190 6 0.250 7 0.290 The equipment we used are • 20mM Tris Buffer pH 8.5 • 33mM MgCl2 • Alkaline Phosphatase (2mg/ml) in 20mM Tris Buffer pH 8.5 • 4mM Glucose-1-phosphate • Acid Molybdate pH 5.0 • Reducing Agent • Distilled Water • Glass Test tubes • Tube Rack • Cuvette • Pipettes and Tips • Water bath set to 37oC The method we used is Method/Protocol: 1. Read the protocol in its entirety before starting. Take note of any additional information that appears in subsequent steps that may influence how previous steps are performed. 2. Using glass tubes, generate the reactions mixtures…The protein calcineurin binds to the protein calmodulin with an association rate of 8.9 × 103 M−1 s −1 and an overall dissociation constant, Kd, of 10 nM. Calculate the dissociation rate, kd , including appropriate units.pH 7.54 PaCO2 28 mm Hg PaO2 45 mm Hg HCO3- 22 mEq/L. Please provide an IPAP, EPAP and FiO2 in ( values not words) Desired FiO2 = [PaO2 (desired) x FiO2 (known)] PaO2 (known)
- © Macmillan Learning You need to prepare an acetate buffer of pH 5.54 from a 0.672 M acetic acid solution and a 2.90 M KOH solution. If you have 780 mL of the acetic acid solution, how many milliliters of the KOH solution do you need to add to make a buffer of pl 5.54? The pKa of acetic acid is 4.76. Be sure to use appropriate significant figures. mL1 Gren: molanty of CsP stlutan -010M = 0.10mol/L vdm me of SoutoL molar mass of csp= 249.68 g/ mole %3D moles af CSP = molahty of CSPX volume of shreoon mdes of asp = 0.10 mol/xIL moles of cSp: 0-10 mol mass of csP-moles of CSPXmolar mass of CsP = 0.10mol X249.c689/mde mass of CSP= 24.9689In a 0.1000 M acetic acid solution at 25 degrees celsius , the acid ionizes to the extent of about 1.34 %. Since each molecule of acetic acid which ionizes produces 1 H+ ion and 1 C2H3O2- ion, the concentration in the solution are: HC2H3O2 < -----------> H+ + C2H3O2-